kosovo dave said:
I'm not sure I follow. I think the only other distribution we've learned so far is the geometric distribution but that doesn't seem appropriate here?
The answer depends on exactly you interpret the question---that is, the meaning of "some district had more than 1 attack". The binomial distribution would be OK if you interpret that as asking about any specific, but un-named district, because each attack either lands in that district or not. I suspect that most people would view that interpretation as somewhat artificial.
The alternative interpretation would be that "at least one district suffers more than one attack", and in that case looking at it as the complement of "every district has exactly one attack" would be the easiest way, as suggested in Post #7.
This would have an easy solution if you use the so-called "Multinomial" distribution, but you say you have not seen that yet. However, you can also do it using the ordinary Binomial distribution and conditional probabilities: if we let ##X_i = ## number of hits in district ##i## for ##i = 1,2, \ldots, 12##, then we want to know ##P(E)##, where ##E## is the event
$$E = \{ (X_1,X_2, \ldots, X_{12}) = (1,1, \ldots, 1) \}$$
Note that ##P(E) = P(X_1 = 1) P(E|X_1 = 1)##.
We get ##P(X_1 = 1)## from the Binomial distribution. The conditional probability ##P(E| X_1 = 1)## is just the probability that all 11 remaining districts have one hit each in 11 attacks, so is a smaller version of the original problem, and can be similarly tackled by computing ##P(X_2 = 1 | X_1 = 1)## and ##P(E | X_1 = 1 \: \& \: X_2 = 1)##, etc.